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Consider a triangle $ABC$. Let the angle bisector of angle $A$ intersect side $BC$ at a point $D$ between $B$ and $C$. The angle bisector theorem states that the ratio of the length of the line segment $BD$ to the length of segment $DC$ is equal to the ratio of the length of side AB to the length of side $AC$:

${\displaystyle {\dfrac {|BD|}{|DC|}}={\frac {|AB|}{|AC|}},} {{\dfrac {|BD|}{|DC|}}}={{\dfrac {|AB|}{|AC|}}}$,The vector $A,B,C$ of triangle $ABC$ have respectively position vectors $\vec{a}, \vec{b}, \vec{c}$ with respect to a given origin $O$. Show that the point $D$ where the bisector of $\angle A$ meets $BC$ has position vector $\vec{d}=\frac{\beta \vec b+\gamma \vec c}{\beta+\gamma}$, where $\beta=|\vec c-\vec a|$ and $\gamma=|\vec a-\vec b|$. Hence deduce that the incentre $I$ has position vector $\frac{\alpha \vec a+\beta \vec b+\gamma \vec c}{\alpha+\beta+\gamma}$, where $\alpha=|\vec b-\vec c|$

Added:

We can prove this using the result "The angle of bisector of the angle $\angle A$" of a triangle $ABC$ divides the sides $BC$ in the ration $AB:AC$.

Is there any other way to prove this result?

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Since the angle bisector of $\widehat{BAC}$ is the locus of points $P$ for which $d(P,AB)=d(P,AC)$, the trilinear coordinates of the incenter $I$ are $[1;1;1]$. It follows that the barycentric coordinates of the incenter are $[a;b;c]$, hence the exact barycentric coordinates are $\left[\frac{a}{a+b+c};\frac{b}{a+b+c};\frac{c}{a+b+c}\right]$, i.e.

$$ I = \frac{aA+bB+cC}{a+b+c} $$ as wanted.

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A simple (although possibly tedious) approach is to prove it algebraically. I will simply write the lowercase letters for the vectors.

From the angle equality

$$ \frac{(d-a)\cdot (c-a)}{|c-a|}=\frac{(b-a) \cdot (d-a)}{|b-a|}$$

Since $ d-c$ and $ b-d$ are collinear,

$$ d-c = \delta(b-d) $$ $$ d = \frac{\delta b+ c}{1+\delta}$$ (this is a general result for a point lying on a line determined by two position vectors)

$$ d-a = \frac{\delta (b-a)+(c-a)}{1+\delta}$$

$$ \frac{(d-a).(c-a)}{|c-a|} = \frac{\delta (b-a)\cdot (c-a)}{(1+\delta)|c-a|}+\frac{|c-a|}{1+\delta}$$

$$ \frac{(b-a).(d-a)}{|b-a|} = \frac{\delta |b-a|}{(1+\delta)}+\frac{(c-a)\cdot (b-a)}{(1+\delta)|b-a|}$$

Equate and solve for $\delta$ to get the first part.

For the second part, we use a similar approach. For any two angle bisectors, the incentre will have the general form:

$$ \frac{a+\mu d}{1+\mu} = \frac{b+\eta e}{1+\eta}$$

We get two equations to solve for two variables. This will give the result.

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